Proceeding in the same way we did in the previous example:
1. Rewrite the piecewise function as a sum of Heaviside functions
we have:
f_{0}(t)=0,\quad f_{1}(t)=\cos\left(t+\frac{\pi}{3}\right)
and
\displaystyle t_{1}=\frac{\pi}{3}
Hence, f may be rewritten using the unit step function:
\begin{aligned} & f(t)=f_{0}(t)+u(t-t_{1})(f_{1}(t)-f_{0}(t)) \\ & f(t)=0+u\left(t-\frac{\pi}{3}\right)\left(\cos\left(t+\frac{\pi}{3}\right)-0\right)\end{aligned}
which yields:
f(t)=u\left(t-\frac{\pi}{3}\right)\cos\left(t+\frac{\pi}{3}\right)
2. Apply the Laplace transform
We have:
\mathscr{L}\{f(t)\}=\mathscr{L}\left\{u\left(t-\frac{\pi}{3}\right)\cos\left(t+\frac{\pi}{3}\right) \right\}
Before we evaluate the Laplace transform above, note that, from line 10 of the table:
\displaystyle\mathscr{L}\{g(t-a)u(t-a)\}=e^{-as}G(s)
and comparing with the expression of f, we get:
\displaystyle a=\frac{\pi}{3}
and the cosine function can be rewritten as follows
\cos\left(t+\frac{\pi}{3}\right)=\cos\left(t+\frac{2\pi}{3}-\frac{\pi}{3}\right)
which yields:
g(t)=\cos\left(t+\frac{2\pi}{3}\right)
3. Laplace transform of cos (t+2pi/3)
We have:
G(s)=\mathscr{L}\{g(t)\}=\mathscr{L}\left\{\cos\left(t+\frac{2\pi}{3}\right)\right\}
Using the formula:
\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)
we are able to rewrite g(t) as follows:
\begin{aligned} \cos\left(t+\frac{2\pi}{3}\right) & =\cos(t)\cos\left(\frac{2\pi}{3}\right)-\sin(t)\sin\left(\frac{2\pi}{3}\right) \\ & =-\frac{1}{2}\cos(t)-\frac{\sqrt{3}}{2}\sin(t) \end{aligned}
Thus:
G(s)=\mathscr{L}\left\{\cos\left(t+\frac{2\pi}{3}\right)\right\}=-\frac{1}{2}\cdot\mathscr{L}\{\cos(t)\}-\frac{\sqrt{3}}{2}\cdot\mathscr{L}\{\sin(t)\}
For b=1 in lines 3 and 4 of the transform table, we have then:
\begin{aligned} G(s) & =-\frac{1}{2}\cdot\frac{s}{s^{2}+1^{2}}-\frac{\sqrt{3}}{2}\cdot\frac{1}{s^{2}+1^{2}} \\ & =-\frac{1}{2}\cdot\left(\frac{s+\sqrt{3}}{s^{2}+1}\right) \end{aligned}
4. Result
Therefore,
\begin{aligned}\mathscr{L}\{f(t)\} &=e^{-\frac{\pi}{3}s}G(s) \\ \mathscr{L}\{f(t)\} &=-\frac{e^{-\frac{\pi}{3}s}}{2}\left(\frac{s+\sqrt{3}}{s^{2}+1}\right) \end{aligned}
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